{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

P [Preliminaries]

# P [Preliminaries] - INSTRUCTOR'S SOLUTIONS MANUAL SECTION...

This preview shows pages 1–2. Sign up to view the full content.

INSTRUCTOR’S SOLUTIONS MANUAL SECTION P.1 (PAGE 10) CHAPTER P. PRELIMINARIES Section P.1 Real Numbers and the Real Line (page 10) 1. 2 9 = 0 . 22222222 · · · = 0 . 2 2. 1 11 = 0 . 09090909 · · · = 0 . 09 3. If x = 0 . 121212 · · · , then 100 x = 12 . 121212 · · · = 12 + x . Thus 99 x = 12 and x = 12 / 99 = 4 / 33. 4. If x = 3 . 277777 · · · , then 10 x 32 = 0 . 77777 · · · and 100 x 320 = 7 + ( 10 x 32 ) , or 90 x = 295. Thus x = 295 / 90 = 59 / 18. 5. 1 / 7 = 0 . 142857142857 · · · = 0 . 142857 2 / 7 = 0 . 285714285714 · · · = 0 . 285714 3 / 7 = 0 . 428571428571 · · · = 0 . 428571 4 / 7 = 0 . 571428571428 · · · = 0 . 571428 note the same cyclic order of the repeating digits 5 / 7 = 0 . 714285714285 · · · = 0 . 714285 6 / 7 = 0 . 857142857142 · · · = 0 . 857142 6. Two different decimal expansions can represent the same number. For instance, both 0 . 999999 · · · = 0 . 9 and 1 . 000000 · · · = 1 . 0 represent the number 1. 7. x 0 and x 5 define the interval [0 , 5]. 8. x < 2 and x ≥ − 3 define the interval [ 3 , 2 ) . 9. x > 5 or x < 6 defines the union ( −∞ , 6 ) ( 5 , ) . 10. x ≤ − 1 defines the interval ( −∞ , 1]. 11. x > 2 defines the interval ( 2 , ) . 12. x < 4 or x 2 defines the interval ( −∞ , ) , that is, the whole real line. 13. If 2 x > 4, then x < 2. Solution: ( −∞ , 2 ) 14. If 3 x + 5 8, then 3 x 8 5 3 and x 1. Solution: ( −∞ , 1] 15. If 5 x 3 7 3 x , then 8 x 10 and x 5 / 4. Solution: ( −∞ , 5 / 4] 16. If 6 x 4 3 x 4 2 , then 6 x 6 x 8. Thus 14 7 x and x 2. Solution: ( −∞ , 2] 17. If 3 ( 2 x ) < 2 ( 3 + x ) , then 0 < 5 x and x > 0. Solution: ( 0 , ) 18. If x 2 < 9, then | x | < 3 and 3 < x < 3. Solution: ( 3 , 3 ) 19. Given: 1 /( 2 x ) < 3. CASE I. If x < 2, then 1 < 3 ( 2 x ) = 6 3 x , so 3 x < 5 and x < 5 / 3. This case has solutions x < 5 / 3. CASE II. If x > 2, then 1 > 3 ( 2 x ) = 6 3 x , so 3 x > 5 and x > 5 / 3. This case has solutions x > 2. Solution: ( −∞ , 5 / 3 ) ( 2 , ) . 20. Given: ( x + 1 )/ x 2. CASE I. If x > 0, then x + 1 2 x , so x 1. CASE II. If x < 0, then x + 1 2 x , so x 1. (not possible) Solution: ( 0 , 1]. 21. Given: x 2 2 x 0. Then x ( x 2 ) 0. This is only possible if x 0 and x 2. Solution: [0 , 2]. 22. Given 6 x 2 5 x ≤ − 1, then ( 2 x 1 )( 3 x 1 ) 0, so either x 1 / 2 and x 1 / 3, or x 1 / 3 and x 1 / 2. The latter combination is not possible. The solution set is [1 / 3 , 1 / 2]. 23. Given x 3 > 4 x , we have x ( x 2 4 ) > 0. This is possible if x < 0 and x 2 < 4, or if x > 0 and x 2 > 4. The possibilities are, therefore, 2 < x < 0 or 2 < x < . Solution: ( 2 , 0 ) ( 2 , ) . 24. Given x 2 x 2, then x 2 x 2 0 so ( x 2 )( x + 1 ) 0. This is possible if x 2 and x ≥ − 1 or if x 2 and x ≤ − 1. The latter situation is not possible. The solution set is [ 1 , 2]. 25. Given: x 2 1 + 4 x . CASE I. If x > 0, then x 2 2 x + 8, so that x 2 2 x 8 0, or ( x 4 )( x + 2 ) 0. This is possible for x > 0 only if x 4. CASE II. If x < 0, then we must have ( x 4 )( x + 2 ) 0, which is possible for x < 0 only if x ≥ − 2. Solution: [ 2 , 0 ) [4 , ) . 26. Given: 3 x 1 < 2 x + 1 . CASE I. If x > 1 then ( x 1 )( x + 1 ) > 0, so that 3 ( x + 1 ) < 2 ( x 1 ) . Thus x < 5. There are no solutions in this case.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}